Let the CES utility function be $$U(x,y)=\frac{x^\delta}{\delta}+\frac{y^\delta}{\delta},$$where $\delta\leq1,\delta\neq0.$ For $\delta=-\infty$, the given function is equivalent to $U(x,y)=\min(x,y)$.
How for $\delta=-\infty$ the function becomes $U(x,y)=\min(x,y)$?
Not this CES representation. Use this one (same preferences but different utilities): $$ U(x,y)=\left(x^\delta+y^\delta\right)^\frac{1}{\delta}$$ Now you can (and should) prove that: $$ \lim_{\delta\searrow-\infty}\dfrac{ U(x,y)}{\min(x,y)} = 1. $$
Assume $x=\min(x,y)\le y$ if the minimum is $y$ then just change the labels.
\begin{align*} \dfrac{ U(x,y)}{\min(x,y)} = \dfrac{\left(x^\delta+y^\delta\right)^\frac{1}{\delta}}{\min(x,y)}=\left( \dfrac{x^\delta+y^\delta}{\min(x,y)^\delta}\right)^\frac{1}{\delta}=\left( \dfrac{x^\delta+y^\delta}{\min(x,y)^\delta}\right)^\frac{1}{\delta}=\\ \left( \dfrac{x^\delta}{\min(x,y)^\delta}+\dfrac{y^\delta}{\min(x,y)^\delta}\right)^\frac{1}{\delta}=\left( \left(\dfrac{x}{\min(x,y)}\right)^\delta+\left(\dfrac{y}{\min(x,y)}\right)^\delta\right)^\frac{1}{\delta}=\left(1+\left(\dfrac{y}{x}\right)^\delta\right)^\frac{1}{\delta} \end{align*} Notice that the limit as $\delta$ goes to minus infinity of the term inside the parenthesis is $1$ if $x<y$ and $2$ if $y=x$ regardless the exponent outside the parenthesis, $\frac{1}{\delta}$ converges to zero as $\delta\rightarrow-\infty$ so you have a positive term being raised to the power of zero in the limit, which is 1 as we wanted to show.